The energy-separation curve for two atoms, a distance, r, apart is: U(r)=−A/r^m+B/r^n
Derive and expression for the stiffness of the bond at the equilibrium spacing, in terms of A, B, m, n, and r0.
S=dF/dr at r=r0:
I don't know what F is, but
dU/dr = Am/r^(m+1) - Bn/r^(n+1)
So, plug in r0 and you have
dU/dr(r0) = Am/r0^(m+1) - Bn/r0^(n+1)
To derive the expression for the stiffness of the bond at the equilibrium spacing, we need to find the derivative of the potential energy function U(r) with respect to the distance r, and then evaluate it at the equilibrium spacing r0.
First, let's find the derivative of U(r) with respect to r:
dU/dr = d(-A/r^m + B/r^n)/dr
To differentiate the terms with respect to r, we need to use the power rule for differentiation. The power rule states that if we have a function of the form f(x) = x^k, then the derivative df/dx is given by df/dx = kx^(k-1).
Applying the power rule to the terms in U(r), we get:
dU/dr = (Am/r^(m+1)) - (Bn/r^(n+1))
Now, we can evaluate this derivative at the equilibrium spacing r0. Let's call the derivative at r=r0 as S (stiffness). So, we have:
S = dU/dr evaluated at r0
S = [(Am/r0^(m+1)) - (Bn/r0^(n+1))] evaluated at r=r0
S = Am/r0^(m+1) - Bn/r0^(n+1)
Therefore, the expression for the stiffness of the bond at the equilibrium spacing r0 is:
S = Am/r0^(m+1) - Bn/r0^(n+1)
This expression represents the stiffness of the bond and is written in terms of constants A, B, exponents m and n, and the equilibrium spacing r0.